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Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW. ENTROPY. 8.1 Spontaneous Change 8.2 Entropy and Disorder 8.3 Changes in Entropy 8.4 Entropy Changes Accompanying Changes in Physical State 8.5 A Molecular Interpretation of Entropy

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Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

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  1. Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW ENTROPY 8.1 Spontaneous Change 8.2 Entropy and Disorder 8.3 Changes in Entropy 8.4 Entropy Changes Accompanying Changes in Physical State 8.5 A Molecular Interpretation of Entropy 8.6 The Equivalence of Statistical and Thermodynamic Entropies 8.7 Standard Molar Entropies 8.8 Standard Reaction Entropies 2012 General Chemistry I

  2. ENTROPY (Sections 8.1-8.8) - The 1st law of thermodynamics says if a reaction takes place, then the total energy of the universe remains unchanged. It cannot be used to predict the directionality of a process. - The natural progression of a system and its surroundings (or “the universe”) is from order to disorder, from organized to random. - A new thermodynamic state function is needed to predict directionality and extent of disorder.

  3. 8.1 Spontaneous Change • Spontaneous change is a change that has a tendency to occur without needing to be driven by an external influence. Heat flow - Spontaneous changes need not be fast: e.g. C(diamond)  C(graphite); Mixing of gases H2(g) + 1/2O2(g)  H2O(l)

  4. 8.2 Entropy and Disorder - Energy and matter tend to disperse in a disorderly fashion. • Entropy, S is defined as a measure of disorder. • The second law of thermodynamics: The entropy of an isolated system increases in any spontaneous change. At constant temperature unit: J·K-1 - Entropy is a state function; the change in entropy of a system is independent of the path between its initial and final states.

  5. 8.3 Changes in Entropy - Thermal disorder: arising from the thermal motion of the molecules - Positional disorder: related to the locations of the molecules • DS for a process with changing temperature: → (CV if V is constant, CP if P is constant)

  6. DS for a reversible, isothermal expansion of an ideal gas

  7. EXAMPLE 8.5 In an experiment, 1.00 mol Ar(g) was compressed suddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston. and in the process its temperature was increased from 20.0 oC to 25.2 oC. What is the change in entropy of the gas? To solve this problem, we consider two reversible stages between initial and final states. Then DS(irrev) = DS(rev 1) + DS(rev 2).

  8. 8.4 Entropy Changes Accompanying Changes in Physical State - Phase transition: solid → liquid, Tf (fusion or melting point) liquid → solid, Tb (boilingpoint) - At the transition temperature (such as Tb), The temperature remains constant as heat is supplied. The transfer of heat is reversible. The heat supplied is equal to the enthalpy change due to the constant pressure (at 1 atm). • Entropy of vaporization, DSvap qrev = DHvap > 0 in all cases

  9. - Standard entropy of vaporization, DSvapo : DSvap at 1 bar Some Standard entropies of vaporization at Tb (Table 8.1)

  10. Trouton’s rule: DSvapo = ~85 J·K-1mol-1 There is approximately the same increase in positional disorder for most liquids when evaporating. - Exceptions; water, methanol, ethanol, ··· due to extensive hydrogen bonding in liquid phases - Standard entropy of fusion, DSfuso > 0 in all cases

  11. Temperature dependence of DSvapo • To determine the entropy of vaporization of water at 25 oC • (not at Tb), we can use an entropy change cycle:

  12. 8.5 A Molecular Interpretation of Entropy • The third law of thermodynamics S → 0 as T → 0 • Boltzmann formula S = kB ln W - statistical entropy: kB = 1.381 × 10-23 J·K-1 = R/NA - W : the number of microstates the number of ways that the atoms or molecules in the sample can be arranged and yet still give rise to the same total energy - When we measure the bulk properties of a system, we are measuring an average taken over the many microstates (ensemble) that the system has occupied during the measurement.

  13. EXAMPLE 8.7 Calculate the entropy of a tiny solid made up of four diatomic molecules of a compound such as carbon monoxide, CO, at T = 0 when (a) the four molecules have formed a perfectly ordered crystal in which all molecules are aligned with their C atoms on the left and (b) the four molecules lie in random orientations, but parallel. (a) 4 CO molecules perfectly ordered: (b) 4 CO in random, but parallel: (c) 1 mol CO in random, but parallel:

  14. - Residual entropy at T = 0, arising from positional disorder 4.6 J·K-1 for the entropy of 1 mol CO < 5.76 J·K-1 Nearly random arrangementdue to a small electric dipole moment - Solid HCl; S ~ 0 at T = 0 due to the bigger dipole moment leading strict head-to-tail arrangement

  15. EXAMPLE 8.8 The entropy of 1.00 mol FClO3(s) at T = 0 is 10.1 J·K-1. Interpret it. 4 orientations possible nearly random arrangement

  16. 8.6 The Equivalence of Statistical and Thermodynamic Entropies = Thermodynamic entropy DS = qrev/T behavior of bulk matter Statistical entropy S = k ln W behavior of molecules - Consider a one-dimensional box, for the statistical entropy, • At T = 0, only the lowest energy level occupied • → W = 1 and S = 0 • At T > 0, W > 1 and S > 0 • When the box length is increased at constant T, • the molecules are distributed across more levels. • → W and S increase.

  17. - W = constant × V - For N molecules, - The change when a sample expands isothermally from V1 to V2 is, V2 = nR ln V1 - By raising the temperature, The molecules have access to larger number of energy levels → W and S increase.

  18. - The equations used to calculate changes in the statistical entropy and the thermal entropy lead to the same result. 1. Both are state functions. Number of microstates depends only on its current state. 2. Both are extensive (dependent on “extent”) properties. 2 × no. of molecules = entropy changes from k ln W to 2k ln W 3. Both increase in a spontaneous change. In any irreversible change, the overall disorder increases → no. of microstates increases. 4. Both increase with temperature. When T increases, more microstates become accessible.

  19. 8.7 Standard Molar Entropies Molar entropy, S(T), can be determined from measurement of Cp at different temperatures. For heating at constant P, CP and CP/T → 0 as T→ 0

  20. Standard molar entropy, Smo is the molar entropy of the pure substance at 1 bar.

  21. Standard molar entropy, Smo : light heavy - Diamond (2.4 J·K-1) vs. lead (64.8 J·K-1): rigid bonds vs. vibrational energy levels - H2 (130.7 J·K-1) vs. N2 (191.6 J·K-1): the greater the mass, the closer energy levels - CaCO3 (92.9 J·K-1) vs. CaO (39.8 J·K-1): large, complex vs. smaller, simpler - In general, Smo: gases >> liquids > solids Related to freedom of movement and disordered state

  22. 8.8 Standard Reaction Entropies • Standard reaction enthalpy, DSo, is the difference between the standard molar entropies of the products and those of the reactants, taking into account their stoichiometric coefficients.

  23. EXAMPLE 8.9 Calculate DSo for N2(g) + 3H2(g) → 2NH3(g) at 25 oC.

  24. Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW GLOBAL CHANGES IN ENTROPY 8.9 The Surroundings 8.10 The Overall Change in Entropy 8.11 Equilibrium GIBBS-FREE ENERGY 8.12 Focusing on the System 8.13 Gibbs Free Energy of Reaction 8.14 The Gibbs Free Energy and Nonexpansion Work 8.15 The Effect of Temperature 8.16 Impact on Biology: Gibbs Free Energy Changes in Biological Systems 2012 General Chemistry I

  25. GLOBAL CHANGES IN ENTROPY(Sections 8.9-8.11) 8.9 The Surroundings - The second law refers to an isolated system (system + surroundings = universe). - Only if the total entropy change is positive will the process be spontaneous.

  26. Sometimes DSsurr can be difficult to compute, but in general it can be obtained from the enthalpy change for the process.

  27. 8.10 The Overall Change in Entropy - To use the entropy to judge the direction of spontaneous change, we must consider the change in the entropy of the system plus the entropy change in the surroundings:

  28. EXAMPLE 8.11 Is the reaction spontaneous? 2 Mg(s) + O2(g) → 2 MgO(s) DSo = -217 J·K-1 DHo = -1202 kJ The reaction is spontaneous

  29. - Spontaneous endothermic (DH>0) reactions: There can still be an overall increase in entropy if the disorder of the system increases enough. Summary

  30. - A process produces maximum work if it takes place reversibly. • Clausius inequality DS = > DS - For an isolated system (universe), q = 0 The entropy of an isolated system cannot decrease. - For two given states of the system,ΔS is a state function (path-independent) but ΔStot is not. (See EXAMPLE 8.12)

  31. EXAMPLE 8.12 Calculate DS, DSsurr, and DStot for (a) the isothermal, reversible expansion and (b) the isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K. Explain any differences between the two paths. (a) Isothermal reversible expansion at 292 K

  32. (b) Isothermal free expansion 292 K

  33. 8.11 Equilibrium • Dynamic equilibrium is one where there is no net tendency to change but microscopic forward and reverse • processes occur at matching rates. Thermal equilibrium: no net flow of energy as heat Mechanical equilibrium: no tendency to expand or contract Chemical equilibrium: no net change in composition at thermodynamic equilibrium

  34. GIBBS FREE ENERGY(Sections 8.12-8.16) 8.12 Focusing on the System - at constant T and P • Gibbs free energy, G G = H - TS - Change in Gibbs free energy - at constant T and P The direction of spontaneous change is the direction of decreasing Gibbs free energy.

  35. at constant T and P DGsys < 0 Spontaneous, irreversible DGsys = 0 Reversible DGsys > 0 Nonspontaneous - the condition for equilibrium, DStot = 0, and DG = 0 at constant T and P

  36. EXAMPLE 8.13 Calculate the change in molar Gibbs free energy, DGm, for the process H2O(s) → H2O(l) at 1 atm and (a) 10 oC; (b) 0 oC. Decide for each temperature whether melting is spontaneous or not. Treat DHfus and DSfus as independent of temperature. (a) At 10 oC, = -0.22 kJ·mol-1 < 0; spontaneous melting

  37. (b) At 0 oC, equilibrium (reversible)

  38. Variation of G with temperature: phase transitions - G decreases as its T is raised at constant P. ; H and S vary little with T, S > 0 G↓ = H – T↑S - Decreasing rate of Gm: vapor >> liquid > solid Sm(vapor) >> Sm(liquid) > Sm(solid) In most cases (opposite), heating leads to melting, then boiling.

  39. In some cases, at certain pressures, G for the liquid may never be lower than those of the other two phases. The liquid phase is unstable and the phase transition is solid vapor (sublimation).

  40. 8.13 Gibbs Free Energy of Reaction • Gibbs free energy of reaction • Standard Gibbs free energy of reaction (standard state: pure form at 1 bar) - DGo is fixed for a given reaction and temperature. - DG depends on the composition of the reaction mixture and so it varies – and might even change sign as the reaction proceeds.

  41. Standard Gibbs free energy of formation, DGfo , isthe standard Gibbs free energy of reaction per mole for the formation of a compound from its elements in their most stable form. - For most stable form of elements, DGfo = 0 E.g. DGfo(I2, s) = 0; DGfo(I2, g) > 0 Examples of most stable forms of elements (Table 8.6)

  42. Some Standard Gibbs free Energies of Formation at 25 oC (kJ mol-1) (Table 8.7)

  43. EXAMPLE 8.14 Calculate the standard Gibbs free energy of formation of HI(g) at 25 oC from its standard molar entropy and standard enthalpy of formation.

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